Let $K$ be a field and $R_1,\dots,R_n$ DVRs of $K$ with $m_i$ the maximal ideal of $R_i$ and $R_i \not\subseteq R_j$ for $j\neq i$ . Define $A=\bigcap_{i=1}^n R_i$. Then $A$ is semilocal with maximal ideals $p_i=m_i\cap A$. Also, $A_{p_i}=R_i$. I know there is a $x_i \in p_i$ such that $x_i \not\in p_j$ for $i \neq j$ and $x_i \not\in p_i^2 A_{p_i}\cap A$.

Question: Why is $p_i=x_iA$?

I know $p_1,\dots,p_n$ are (pairwise) coprime and so for $p_1,\dots,p_{i-1}, p_i^2, p_{i+1},\dots,p_n$ but Matsumura's argument is unclear to me.

Reference: Theorem 12.2 in Matsumura, Commutative Ring Theory.


To prove that $p_i = x_iA$, you can do so locally, at every maximal ideal. Since $A$ is semilocal with maximal ideals $p_j$, and both $p_i, x_iA$ localize to the unit ideal in $A_{p_j}$ for $j \ne i$, it suffices to show that $p_iA_{p_i} = x_iA_{p_i}$. But this is true: $p_iA_{p_i} = m_i = x_iR_i = x_iA_{p_i}$ (as $x_i$ was chosen to be a uniformizing parameter for the DVR $R_i$).


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